# Drill 32 L2 update

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/read/17/tee-c17-s32

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/section/tee-c17-s32/

## Question

**Question.** w = [0.5,−0.2], data gradients = [0.1,0.3], λ = 0.2, η = 0.1. Use(λ/2)Σw².

## Answer — step by step

### Given

Old weights = [0.5,−0.2]; data gradients = [0.1,0.3].

λ = 0.2; η = 0.1; penalty = (λ/2)Σw².

### To find

Find penalty gradients, total gradients and updated weights.

### Formula — which one and why

**L2 penalty with a factor of one-half.** Square each weight, sum and multiply by λ/2. The one-half cancels the factor 2 when differentiating, so the extra gradient is λw.

λ = penalty strength; w = old weight; J\_data and g\_data = data loss and its gradient.



$$
J=J_{\mathrm{data}}+\frac\lambda2\sum_iw_i^2
$$





$$
g_i=g_{\mathrm{data},i}+\lambda w_i
$$





$$
w_i^{\mathrm{new}}=w_i-\eta g_i
$$



**Yaad rakhne ka cue:** Half wala L2 derivative mein lambda × weight deta hai.

### Solution — step by step

### Formula and its reason

The derivative of (λ/2)w² is λw: the 2 from the square cancels the half. Add that regularization gradient to the given data gradient before updating.



$$
g_{j,\mathrm{total}}=g_{j,\mathrm{data}}+\lambda w_j
$$





$$
w_{j,\mathrm{new}}=w_j-\eta g_{j,\mathrm{total}}
$$



**Step 1: Weight 1.**



$$
\lambda w_1=0.2\times0.5=0.1
$$





$$
g_{1,\mathrm{total}}=0.1+0.1=0.2
$$





$$
w_{1,\mathrm{new}}=0.5-0.1(0.2)=0.48
$$



**Step 2: Weight 2.**



$$
\lambda w_2=0.2\times(-0.2)=-0.04
$$





$$
g_{2,\mathrm{total}}=0.3+(-0.04)=0.26
$$





$$
w_{2,\mathrm{new}}=-0.2-0.1(0.26)=-0.226
$$



**Final answer: [0.48, −0.226].** The second data gradient dominates the regularization term, explaining why that negative weight becomes more negative.